9k^2-19k=-9k+18

Solution for 9k^2-19k=-9k+18 equation:

9k^2-19k=-9k+18

We move all terms to the left:

9k^2-19k-(-9k+18)=0

We get rid of parentheses

9k^2-19k+9k-18=0

We add all the numbers together, and all the variables

9k^2-10k-18=0

a = 9; b = -10; c = -18;
Δ = b2-4ac
Δ = -102-4·9·(-18)
Δ = 748
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{748}=\sqrt{4*187}=\sqrt{4}*\sqrt{187}=2\sqrt{187}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{187}}{2*9}=\frac{10-2\sqrt{187}}{18}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{187}}{2*9}=\frac{10+2\sqrt{187}}{18}
$